Denote $ \sigma$ the bijection on $ \{1,...,5\}$ given by $ \sigma(i) = 6 - i$, denote $ S = \{n\in\mathbb{N}|n > 33333\text{ and }n\text{ is a digit permutation of }12345\}$. Denote the numbers in $ S$ as $ [a,b,c,d,e]$ and consider the differences \[ {T_{abcde} = [a,b,c,d,e]^2 - [a_{\sigma},b_{\sigma},c_{\sigma},d_{\sigma},e_{\sigma}]^2 = 66666\cdot [(a - a_\sigma),(b - b_\sigma),(c - c_\sigma),(d - d_\sigma),(e - e_\sigma)].} \] Then there is to prove that $ S = S_1\overset\cdot{\smash\cup} S_2$ with $ \sum_{[a,b,c,d,e]\in S_1}T_{[a,b,c,d,e]} = \sum_{[a,b,c,d,e]\in S_2}T_{[a,b,c,d,e]}$. Now note that $ x - x_\sigma\in\{ - 4, - 2,0,2,4\}$, and consider the differences $ T_{[p,q,r,s,t]} - T_{[p_\tau,q_\tau,r_\tau,s_\tau,t_\tau]}$, where $ \tau = (12)(3)(4)(5)$, with $ 2$ appearing left of $ 1$ in $ [p,q,r,s,t]\in S$ (note $ p_\tau = p$) -- and note that each element of $ S$ is uniquely represented in this way. The differences are one of $ 9, 90, 99, 900, 990, 999$, appearing respectively $ 6,6,6,4,4,4$ times. Since $ 999 - 990 + 9 = 0$, $ 999 - 900 + 99 = 0$, $ 990 - 900 + 90 = 0$ and $ 99 - 90 - 9 = 0$, we can indeed split $ S$ in two disjoint sets with equal $ T$-sums, and we're done.